How do you solve #1/x+1/2=(5x)/6+1/3#?

Answer 1

#x = 6/5" or " x = -1#

We could treat this as a question on algebraic fractions and add them, but as we are going to solve the equation we don't really want the fractions at all.

Multiply by the LCM of the denominators to cancel them completely. #LCM = color(red)(6x)#
#(color(red)(6cancelx) xx1)/cancelx +(color(red)(cancel6^3x) xx 1)/cancel2 = (color(red)(cancel6x) xx5x)/cancel6 +(color(red)(cancel6^2x) xx1)/cancel3#
#6+3x= 5x^2 +2x" "larr# a quadratic, make #=0#
#5x^2+2x-3x-6 = 0#
#5x^2 -x-6=0#
#(5x-6)(x+1)=0#

Set each factor equal to 0.

#5x -6 =0 " "rarr x = 6/5#
#x+1 = 0" "rarr x =-1#
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Answer 2

To solve the equation 1/x + 1/2 = (5x)/6 + 1/3, you can follow these steps:

  1. Find a common denominator for the fractions on both sides of the equation.
  2. Multiply each term by the common denominator to clear the fractions.
  3. Simplify the resulting equation.
  4. Solve for the variable x.
  5. Check the solution by substituting it back into the original equation.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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